Basic notions of probability theory

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1 Basic notions of probability theory

2 Contents o Boolean Logic o Definitions of probability o Probability laws

3 Objectives of This Lecture What do we intend for probability in the context of RAM and risk analysis? To be able to use the probability laws for RAM and risk analysis

4 Why a Lecture on Probability? Lecture 1, Slide 13: «Operative definition of reliability: Probability that an item performs its required function, under given environmental and operational conditions for a stated period of time Lecture 1, Slide 28:

5 Basic Definitions

6 Definitions: experiment, sample space, event xperiment ε: process whose outcome is a priori unknown to the analyst (all possible outcomes are a priori known) Sample space Ω: the set of all possible outcomes of ε. vent : a set of possible outcomes of the experiment ε (a subset of Ω): the event occurs when the outcome of the experiment ε is one of the elements of.

7 Boolean Logic

8 Definition: Certain events Boolean Logic Logic of certainty: an event can either occur or not occur Indicator variable X = 0, when does not occur 1, when occurs

9 Certain vents (xample) (ε = die toss;ω={1,2,3,4,5,6};=odd number) I perform the experiment and the outcome is 3 vent, X True X =1

10 Boolean Logic Operations Negation: Union: X 1 1 X 1 X 1 1 X AB A B j j A, B C X X X X X j A, B j A B A B Intersection: X X X AB A B

11 Boolean Logic Operations Negation: Union: X 1X X 1 1 X 1 X 1 1 X AB A B j j A, B C X X X X X j A, B j A B A B Intersection: X X X A B A B 0 Definition: A and B are mutually exclusive events if X A B

12 xercise 1

13 xercise 1 In an energy production plant there are two pumps: pump P 1 and pump P 2. ach pump can be in three different states: 0 = operating 1 = degraded 2 = failed Consider the following events: A = pump P 1 is degraded B = both pump P 1 and pump P 2 are failed C = pump P 2 is not failed Questions: Which is the sample space Ω of the state of the pumps? Represent it graphycally? Represent events A, B and C Find and represent A C, A U B, (A C) U B

14 Question 1: Sample space Ω? Which is the sample space Ω? Ω={ 00, 01, 02, 10, 11, 12, 20, 21, 22 } Pump Pump 1

15 Question 2 A= pump 1 is degraded Pump Pump 1

16 Question 2 A= pump 1 is degraded Pump Pump 1

17 Question 2 B = both pump 1 and pump 2 are failed Pump Pump 1

18 Question 2 B = both pump 1 and pump 2 are failed Pump Pump 1

19 Question 2 C pump 2 is not failed Pump Pump 1

20 Question 2 C pump 2 is not failed Pump Pump 1

21 Question 3 A C A C

22 Question 3 A C A C Pump Pump 1

23 Question 3 A U B A B

24 Question 3 A U B A B Pump Pump 1

25 Question 3 (A C) U B A B C

26 Question 3 (A C) U B A B C Pump Pump 1

27 Uncertain vents

28 Uncertain vents Let us consider: the experiment ε, its sample space Ω and the event. (ε = die toss; Ω={1,2,3,4,5,6}; =Odd number) vent, X? True X =1 False X =0

29 Uncertain vents: Let us consider: the experiment ε, its sample space Ω, event. vent, X? True X =1 False X =0 Uncertain events can be compared probability of = p() Probability for comparing the likelihood of events Money for comparing the value of objects

30 Probability theory

31 Probability theory: Kolmogorov Axioms 1. 0 p 1 2. p p Addition law: Let 1,, n be a finite set of mutually exclusive events: (X i X j = ). n n pui p i1 i1 i 1 2

32 Definitions of probability

33 Definitions of probability

34 Three definitions of probability 1. Classical definition 2. mpirical Frequentist Definition 3. Subjective definition

35 1. Classical Definition of Probability Let us consider an experiment with N possible elementary, mutually exclusive and equally probable outcomes: A 1, A 2,.,A N and the event: A 1 A... 2 A M p = number of outcomes resulting in total number of possible outcomes = M N

36 1. Classical Definition of Probability Let us consider an experiment with N possible elementary, mutually exclusive and equally probable outcomes: A 1, A 2,.,A N : A 1 A... 2 A M When is it applicable? Gambling (e.g. tossing of a die) If no evidence favouring one outcome over others p = number of outcomes resulting in total number of possible outcomes = M N

37 1. Classical Definition of Probability (criticisms) Let us consider an experiment with N possible elementary, mutually exclusive and equally probable outcomes: A 1, A 2,.,A N : When is this requirement met? A 1 A... 2 A M In most real life situations the outcomes are not equally probable! p = number of outcomes resulting in total number of possible outcomes = M N

38 2. Frequentist Definition of Probability Let us consider: the experiment ε, its sample space Ω and an event. ε = die toss;ω={1,2,3,4,5,6};={odd number} - n times ε, occurs k times (n = 100 die tosses k=48 odd numbers) - k/n = the relative frequency of occurrence of (k/n = 48/100=0.48) k lim n n = p p is defined as the probability of

39 2. Frequentist Definition of Probability (criticisms) lim n k n = p This is not a limit from the matemathical point of view [limit of a numerical series] We tacitly assume that the limit exists It is not possible to repeat an experiment an infinite number of times Possible way out? probability as a physical characteristic of the object: the physical characteristics of a coin (weight, center of mass, ) are such that when tossing a coin over and over again the fraction of head will be p

40 2. Frequentist Definition of Probability (criticisms) lim n k n = p Applicable only to those events for which we can conceive of a repeatable experiment (e.g. not to the event «Next Tuesday: Juventus Porto = 2-0») The experiment conditions cannot be identical let us consider the probability that a specific valve V of a specific Oil & Gas plant will fail during the next year what should be the population of similar valves? Large population: all the valves used in industrial plants. Considering data from past years, we will have a large number n, but data may include valves very different to V Small population: valve used in Oil&Gas of the same type, made by the same manufacturer with the same technical characteristics too small n for limit computation. Similarity Vs population size dilemma

41 2. Frequentist Definition of Probability (criticisms) k lim n n = p Some events (e.g. in the nuclear industry) have very low probabilities (e.g. p 10-6 ) (RAR VNTS) Very difficult to observe The frequentist definition is not applicable

42 3. Subjective Definition of Probability P() is the degree of belief that a person (assessor) has that will occur, given all the relevant information currently known to that person (background knowledge) Probability is a numerical encoding of the state of knowledge of the assessor (De Finetti: probability is the feeling of the analyst towards the occurrence of the event ) P() is conditional on the background knowledge K of the assessor: P()= P( K) Background knowledge typically includes data/models/expert knowledge If the background knowledge changes the probability may change Two interpretations of subjective probability: Betting interpretation Reference to a standard for uncertainty

43 3. Betting Interpretation P {Iceland will win next FIFA world cup K} = 0.05 Betting agency If Iceland wins next FIFA WORLD CUP, you will receive 1 Bettor (Assessor) I bet 0.05

44 3. Betting Interpretation: two sideness of the bet Betting agency If Iceland win next FIFA WORLD CUP, you will receive 1 Bettor I bet 0.05 Betting agency If Iceland does not win next FIFA WORLD CUP, you will receive 1 Bettor I bet 0.95

45 3. Betting Interpretation P() is the amount of money that the person assigning the probability would be willing to bet if a single unit of payment were given in return in case event were to occur and nothing otherwise. Fair betting: The opposite must also hold: 1- P() is also the amount of money that the person assigning the probability would be willing to bet if a single unit of payment were given in return in case event were not to occur and nothing otherwise.

46 3. Betting Interpretation (Criticism) P() is the amount of money that the person assigning the probability would be willing to bet if a single unit of payment were given in return in case event were to occur and nothing otherwise. The opposite must also hold (1- P() ) is also the amount of money that the person assigning the probability would be willing to bet if a single unit of payment were given in return in case event were not to occur and nothing otherwise. probability assignment depends from the value judgment about money and event consequences (the assessor may even think that in case of LOCA in a Nuclear Power Plant he/she will die and so the payment will be useless)

47 3. Reference to a standard for uncertainty P( K) is the number such that the uncertainty about the occurrence of is considered equivalent by the person assigning the probability (assessor) to the uncertainty about drawing a red ball from an urn of 100 balls containing P( K)*100 red balls. ={Germany will win next FIFA WORLD CUP} P( K)=0.33 urn

3. Reference to a standard for uncertainty P() is the number such that the uncertainty about the occurrence of is considered equivalent by the person assigning

48 3. Reference to a standard for uncertainty P() is the number such that the uncertainty about the occurrence of is considered equivalent by the person assigning the probability (assessor) to the uncertainty about drawing a red ball from an urn containing P()*100 red balls ={Germany will win next FIFA WORLD CUP} P()=0.33 urn

49 NURG 75/014 Nureg 75/014: Reactor Safety Study: An Assessment of Accident Risks in U.S. Commercial Nuclear Power Plants (Rasmussen et al., 1975) «The overall probability of a complete core meltdown is about per reactor per year»

50 NURG 75/014: probability interpretation Nureg 75/014: Reactor Safety Study: An Assessment of Accident Risks in U.S. Commercial Nuclear Power Plants (Rasmussen et al., 1975) «The overall probability of a complete core meltdown is about per reactor per year Interpretation? the likelihood of an average citizen s being killed in a reactor accident is about the same as the chance «Widely quoted and of being killed by a falling meteorite much criticized statement»

51 Probability laws

52 Probability laws (1) Union of two non-mutually exclusive events A B P A B = *

53 Probability laws (1) Union of two non-mutually exclusive events A A B B P A B =P A +P B P A B It can be demonstrated by using the three Kolmogorov axioms* P A B P A +P B Rare event approximation: A and B events are considered as mutually exclusive (A B = ) P A B = 0 P A B =P A +P B *

54 Probability laws (2) Union of non-mutually exclusive events: = ራ i=1,,n i 3 1 2

P( ) n P( U j j1 n U ) j1 P( j ) ) n1 n i1 ji1 P( i j ) Rare event

55 Probability laws (2) Union of non-mutually exclusive events: = ራ i=1,,n i n n 1 n P = P i P i j n+1 P 1 2 n i=1 i=1 j=i+1 Upper bound Lower bound P( P( ) n P( U j j1 n U ) j1 P( j ) ) n1 n i1 ji1 P( i j ) Rare event approximation: events are considered as mutually exclusive ( i j =, i, j, i j) P = σ n i=1 P i

56 o Conditional Probability of A given B o vent A is said to be statistically independent from event B if: o If A and B are statistically independent then: Probability laws (3) A B A B ) ( ) ( ) ( B P B A P B A P ) ( ) ( A P B A P ) ( ) ( ) ( B P A P B A P

57 xercise 2 P a and P b are two pumps in parallel sharing a common load. Let A denoting the event pump P a is failed and B the event pump P b is failed with P(A) = 0.040, P(B) = and P(A U B) = Questions: 1. What is the probability that both pumps are failed? 2. What is the probability tha pump P a is also failed given that pump P b is failed? 3. What is the probability tha pump P b is also failed given that pump P a is failed?

58 x. 2 - Question 1 - Solution The probability that both pumps are failed: A P(A U B) = P(A) + P(B) - P(A B) B A B P(A B) = P(A) + P(B) - P(A U B) = = 0.035

59 x. 2: Other questions - Solutions The probability that pump P a is also failed given that pump P b is failed: A P(A B)=P(A B)/P(B)= =0.035/0.080= 0.46 A B B The probability that pump P b is also failed given that pump P a is failed: P(B A) = P(A B)/P(A) = = 0.035/0.040 = 0.875

60 Theorem of Total Probability Let us consider a partition of the sample space into n mutually exclusive and exhaustive events. In terms of Boolean events: i j 0 i j n j1 j

61 Theorem of Total Probability Let us consider a partition of the sample space into n mutually exclusive and exhaustive events. In terms of Boolean events: i j 0 i j n j1 j A Given any event A in, P(A)=

62 Theorem of Total Probability Let us consider a partition of the sample space into n mutually exclusive and exhaustive events. In terms of Boolean events: i j 0 i j n j1 j A Given any event A in, its probability can be computed from P j, P A j with j = 1,, N P( A) P( A ) P( 1) P( A 2) P( 2 )... P( A ) P( 1 n n ) Demonstration: A= j A j P A = σn j=1 P A j = σn j=1 P A j P( j )

63 xercise 3 A motor operated valve opens and closes intermittently on demand to control the coolant level in an industrial process. An auxiliary battery pack is used to provide power for approximately the 0.5% of the time when there are plant power outages. The demand failure probability of the valve is when operated from the plant power and when operated from the battery pack. You are requested to: find the demand failure probability assuming that the number of demands is independent of the power source. Is the increase due to the battery pack operation significant?

64 xercise 3: Solution A motor operated valve opens and closes intermittently on demand to control the coolant level in an industrial process. An auxiliary battery pack is used to provide power for approximately the 0.5% of the time when there are plant power outages. The demand failure probability of the valve is when operated from the plant power and when operated from the battery pack. You are requested to: find the demand failure probability assuming that the number of demands is independent of the power source. Is the increase due to the battery pack operation significant? X = {power outage} P X = 0.005, P തX = 1 P X = Y = {valve failure} P Y തX = P Y X = Theorem of Total Probability: P Y = P Y X P X + P Y തX P തX = = The battery pack operation is causing a 15% increase of the valve failure probability, even if the battery is used only for the 0,5% of the time

65 Bayes Theorem Let us consider a partition of the sample space into n mutually exclusive and exhaustive events j. We know: P( j ), j = 1,, n

66 Bayes Theorem vent A has occurred n j j j i i i i i P A P P A P A P P A P A P 1 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( Let us consider a partition of the sample space into n mutually exclusive and exhaustive events j. We know: P( j ), j = 1,, n Can I use this information to update the probability of P j? A

67 Bayes Theorem vent A has occurred n j j j i i i i i P A P P A P A P P A P A P 1 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( theorem of total probability P( i A) P(A) Let us consider a partition of the sample space into n mutually exclusive and exhaustive events j. We know: P( j ), j = 1,, n Can I use this information to update the probability of P j? A

68 xercise 4 Consider a pile foundation, in which pile groups are used to support the individual column footings. ach of the pile group is designed to support a load of 200 tons. Under normal condition, this is quite safe. However, on rare occasions the load may reach as high as 300 tons. The foundation engineer wishes to know the probability that a pile group can carry this extreme load of up to 300 tons. Based on experience with similar pile foundations, supplemented with blow counts and soil tests, the engineer estimated a probability of 0.70 that any pile group can support a 300-ton load. Also, among those that have capacity less than 300 tons, 50% failed at loads less than 280 tons. To improve the estimated probability, the foundation engineer ordered one pile group to be proof-loaded to 280 tons. If the pile group survives the specified proof load, what is the probability that the pile group can support a load of 300 tons?

69 xercise 4: Solution B = event that the capacity of pile group 300 tons T = event of a successful proof load to 280 tons. P B = 0,7 P തT തB = 0.5 P T B = 1 Bayes theorem then gives P B T = P T B P(B) σ C P T C P(C) = P T B P(B) P T B P B + P T തB P തB = ,7 + (1 0.5) 0.3 If the proof test is successful, the required probability is increased from 0.70 to

70 The Bayesian Subjective Probability Framework P( K) is the degree of belief of the assigner with regard to the occurrence of (numerical encoding of the state of knowledge K - of the assessor) Bayes Theorem to update the probability assignment in light of new information Updated P i A, K = σn j=1 P(A i, K) P i K P(A j, K) P j K Old new information

Basic notions of probability theory

Basic notions of probability theory Contents o Boolean Logic o Definitions of probability o Probability laws Why a Lecture on Probability? Lecture 1, Slide 22: Basic Definitions Definitions: experiment,